Re: infinity

In article <MPG.1d7f9000c6e111a298a19f@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:

> > > > I mapped paths from the same tree to sets of naturals so as to
> > > > establish a bijection between them and the power set of N.
> > >
> > > Now you are trying to change things, but fine, let's examine what you
> > > are now proposing. You claim to have branches mapped to all the
> > > naturals in a 1-1 correspondence, and therefore countable. Then you
> > > claim that the set of paths is the powerset of that set, and
> > > therefore uncountable? You really should draw pictures before writing
> > > words. Then you won't waste 1,000 of them sounding stupid. Here's
> > > your tree, with each number being a branch/node:
> > >
> > > .
> > > 1 2
> > > 3 4 5 6
> > > 7 8 9 10 11 12 13 14
> > >
> > > Which path contains the set {1,2,8,9}? None? How can this be? Is
> > > there really a bijection between the paths of this tree and the
> > > powerset of the naturals? Absolutely not. Each path will start with
> > > either 1 or 2, so one of those will ALWAYS be a member of any path.
> > > For any given subset defined by a path in this tree, the next largest
> > > element to any n in the subset is either 2n+1 or 2n+2, not any
> > > arbitrary m>n. This is obviously not corresponding to the power set
> > > of N, but of a set of subsets which, given proper attention, can
> > > undoubtedly be shown to be exactly N/2, if there are N branches,
> > > since this is the correct answer.
> > > >
> > > > > Using the first, you "proved" that the branches are "countable",
> > > > > and using the second you "proved" that the paths are
> > > > > "uncountable".
> > > >
> > > > Precisely!
> >
> > > Ummmm.....didn't you just deny that you used two different trees, and
> > > claim they were the same, above?
> >
> > No! I used one maximal binary tree, which has both branches and paths. I
> > showed that the branches biject with the naturals and the paths bijecte
> > with the power set of the naturals,
> But look at the tree above. Is this not what you just described, a tree where
> each branch is a natural? If that is the case, then the paths do NOT
> correspond
> to the power set, since most of the subsets of N are not included as paths.
> Did
> you not see my question, above? Which path corresponds to the set {1,2,8,9},
> which would be a member of the power set? None. You have 8 paths with your 14
> nodes in this initial segment. Why do you not have 2^14 paths, if the paths
> biject with the power set? That is quite a discrepancy.

The disrepancy is between what I described and what TO misunderstood me
to have described. If he could actually read, perhaps such discrepancies
would not occur, at least so often.

Each path in an infinite binary tree consists of a sequence of branches
which can be uniquely numbered with the infinitely many finite naturals,
so that all paths have a branch 1 and a ranch 2 and a branch 3 and so on
without end. Each path determines a set by including the numbers for
each right branch and excluding the numbers for each left branch (the
reverse would work equally well).

This clearly bijects paths with subsets of N (the infinite set of finite
naturals).

Each node in the same tree can be matched with a unique natural in
binary notation as follow: The root node is 1. For each left branch on
the finite path from the root node to a given node, append 0, and for
each right branch append 1. So the children of the root will be numbered
10 and 11, the grandchildren 100, 101, 110, and 111, and so on. the
number of binary digits will be one more than the number of branches
between the root and the node which matches the binary number.

This clearly bijects the set of nodes of that same maximal binary tree
with N (the infinite set of finite natural numbers).

So that unless TO can construct a bijection between N (the infinite set
of finite natural numbers) and P(N), there are more (in the sense of
Cantor) paths than nodes.
.

Relevant Pages

  • Re: Orlow cardinality question
    ... >> the infinite set of natural numbers and the sets they each define. ... >>> infinite in a a maximal binary tree. ... > bijected with the naturals. ... I never said the bijection doesn't exist. ...
    (sci.math)
  • Re: infinity
    ... I used one maximal binary tree, which has both branches and paths. ... >>> showed that the branches biject with the naturals and the paths bijecte ... > So that unless TO can construct a bijection between N (the infinite set ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... I consider an infinite tree. ... If you mean that there is a bijection between paths and reals, ... Assume a binary tree in which each node is a parent and has a left-child ...
    (sci.math)
  • Re: infinity
    ... You are constructing a bijection with the elements of the tree and the ... elements of the naturals and of the powerset of the naturals. ... We can also construct a bijection between the paths of the binary tree ...
    (sci.math)
  • Re: Cantor and the binary tree
    ... I use each node in the max binary tree to correspond to a binary natural ... bijection and whether the second defined correspondence, ... paths and sets of naturals, is a bijection, and they both are. ...
    (sci.math)